1. Day 4 Analysis of Qualitative data Anne Segonds-Pichon v2019-03

2. Outline of this section
Comparing 2 groups (one factor)
Fisher’s exact or Chi2 tests
Example: Dancing cats
More than one factor
Logistic regression
Example: Methylation data

3. Qualitative data = not numerical
= values taken = usually names (also nominal)
e.g. causes of death in hospital
Values can be numbers but not numerical
e.g. group number = numerical label but not unit of measurement
Qualitative variable with intrinsic order in their categories = ordinal
Particular case: qualitative variable with 2 categories: binary or dichotomous
e.g. alive/dead or presence/absence
Can be good or horrible data qualitative data do not use many stats, one is chi square
Cats and dogs, first plot it

4. Comparing 2 groups
One factor

5. Fisher’s exact and Chi2 Example: cats.dat Cats trained to line dance
2 different rewards: food or affection
Question: Is there a difference between the rewards?
Is there a significant relationship between the 2 variables?
does the reward significantly affect the likelihood of dancing?
To answer this type of question:
Contingency table
Fisher’s exact or Chi2 tests
Food
Affection
Dance ?
No dance
Is there a difference between the rewards?
Dogs more happy, cats more picky (on demonstration is the other way around, so cats more picky
But first: how many cats do we need?

6. Exercise 14: Power calculation cats.dat
Preliminary results from a pilot study: 25% line-danced after having received affection as a reward vs. 70% after having received food.
How many cats do we need?
power.prop.test(p1= 0.25, p2= 0.7, sig.level= 0.05, power= 0.8)
Providing the effect size observed in the experiment is similar to the one observed in the pilot study, we will need 2 samples of 18 to 19 cats to reach significance (p<0.05) with a Fisher’s exact test.

7. Plot cats data (From raw data)
cats<-read.delim("cats.dat")
head(cats)
plot(cats$Training, cats$Dance, xlab = "Training", ylab = "Dance")
table(cats)

8. Plot cats data (From raw data)
contingency.table <- table(cats)
contingency.table100<-prop.table(contingency.table,1)
contingency.table100<-round(contingency.table100*100)
contingency.table100
barplot(t(contingency.table100), legend.text=TRUE)

9. Plot cats data (From raw data) Prettier
barplot(t(contingency.table100),
col=c("chartreuse3","lemonchiffon2"),
cex.axis=1.2,
cex.names=1.5,
cex.lab=1.5,
ylab = "Percentages",
las=1)
legend("topright",
title="Dancing",
inset=.05,
c("Yes","No"),
horiz=TRUE,
pch=15,
col=c("chartreuse3","lemonchiffon2"))

10. Chi-square and Fisher’s tests
Chi2 test very easy to calculate by hand but Fisher’s very hard
Many software will not perform a Fisher’s test on tables > 2x2
Fisher’s test more accurate than Chi2 test on small samples
Chi2 test more accurate than Fisher’s test on large samples
Chi2 test assumptions:
2x2 table: no expected count <5
Bigger tables: all expected > 1 and no more than 20% < 5
Yates’s continuity correction
All statistical tests work well when their assumptions are met
When not: probability Type 1 error increases
Solution: corrections that increase p-values
Corrections are dangerous: no magic
Probably best to avoid them

11. Chi-square test Example with ‘cats’
In a chi-square test, the observed frequencies for two or more groups are compared with expected frequencies by chance.
With observed frequency = collected data
Example with ‘cats’
Your frequency, compare the frequencies with the expected ones observed by chance. HOW DIFFERENT IS WHAT I observe from what I expect? THIS DIFFERENCE CAN BE BIG SO SIGNOFICANT OR NOT

12. Is 25.35 big enough for the test to be significant?
Chi-square test
Formula for Expected frequency = (row total)*(column total)/grand total
Example: expected frequency of cats line dancing after having received food as a reward:
Expected = (38*76)/200=14.44
Alternatively:
Probability of line dancing: 76/200
Probability of receiving food: 38/200
(76/200)*(38/200)=0.072
Expected: 7.2% of 200 = 14.44
Distance between expected and observed
Chi2 = ( )2/ ( )2/ ( )2 / ( )2/14.4
= 25.35
Is big enough for the test to be significant?

13. Chi-square and Fisher’s Exact tests
Odds of dancing
48/114 = affection
28/10 = food
Ratio of the odds
food
affection
= 6.6
Answer: Training significantly affects the likelihood of cats line dancing (p=4.8e-07).

14. Chi-square and Fisher’s Exact tests
From a matrix
Sometimes data to test are just a few values:
These numbers come from an experiment/query
In the entire genome (n=29395), 221 genes are associated with a function (e.g. antigen binding) . From an experiment, we identified 342 genes out of which 23 are showing such an association.
Question: Is the enrichment significant?
We need to create a matrix with the 4 values:
enrichment <- matrix(c(23, 342, 221, 29395), nrow=2, ncol=2, byrow = TRUE)
chisq.test(enrichment, correct=F) fisher.test(enrichment)
Answer: Yes there is a significant enrichment in our list of genes (p < 2.2e-16).

15. with more than one factor
Comparisons
with more than one factor

16. Real Example: DNA Methylation
Genome
Read 1
Read 2
Read 3
SRR _0021:5:1:1361: u
SRR _0021:5:1:1361: u
SRR _0021:5:1:1355: m
SRR _0021:5:1:1361: u
SRR _0021:5:1:1355: m
SRR _0021:5:1:1358: u
SRR _0021:5:1:1361: u
SRR _0021:5:1:1353: u

17. Chi-square and Fisher’s Exact tests
More than one factor
Still comparing 2 groups: one factor/condition of interest (e.g. treatment vs control)
but each group has 3 biological samples/3 experiments
Example: dna_methylation_format2.csv
Values are independent observations of individual DNA bases.
A DNA base is either methylated (Meth) or unmethylated (Unmeth).
Equivalent to a 2-way ANOVA but with a binary outcome:
Binary logistic regression

18. Logistic regression dna_methylation_format2.csv
For each gene: Is there a difference in methylation between the 2 conditions (AL and DR) accounting for the variability between samples?
Question 1: is the difference between samples significant? (experimental variability)
Question 2: is the difference between conditions significant accounting for the variation between samples?

19. Exercise 15: dna_methylation_format2.csv
Load dna_methylation_format2.csv
Use head() to get to know the structure of the data
Calculate the percentage of methylation for each sample
Plot the percentage for each gene separately and within each gene for each sample.
No clues but here is how the graph should look -ish.

20. Exercise 15: dna_methylation_format2.csv - Answers
## Load the file ##
dna.methylation <- read.csv("dna_methylation_format2.csv")
head(dna.methylation)
## calculate percentage ##
dna.methylation$MethPercent<-(dna.methylation$Meth/(dna.methylation$Meth+dna.methylation$Unmeth)*100)
## create 2 files: one for each gene ##
dna.methylation.Pno1<-dna.methylation[dna.methylation$Gene=="Pno1",]
dna.methylation.Pnpla5<-dna.methylation[dna.methylation$Gene=="Pnpla5",]
head(dna.methylation.Pnpla5)

21. Exercise 15: dna_methylation_format2.csv - Answers
## plot percentage methylation for each sample and for each gene ###
par(mfrow=c(1,2))
## dna.methylation.Pno1<-as.data.frame(dna.methylation.Pno1) ##
barplot(dna.methylation.Pno1$MethPercent,
names.arg = dna.methylation.Pno1$Sample,
col=rep(c("lightblue","lightpink"), each=3),
main = "Pno1",
ylim=c(0,100),
cex.names=0.6,
ylab = "Percentage Methylation",
las=1)
## dna.methylation.Pnpla5<-as.data.frame(dna.methylation.Pnpla5) ##
barplot(dna.methylation.Pnpla5$MethPercent,
names.arg = dna.methylation.Pnpla5$Sample,
main = "Pnpla5",
Ta-dah!

22. Exercise 16: dna_methylation_format2.csv
Question 1: is the difference between samples significant? (experimental variability)
Work a Chi2 test for each group/gene so that you can workout the experimental variability.
Restructure the file in the long format (1 file per gene) for Question 2

23. Exercise 16: dna_methylation_format2.csv - Answers
Question 1: is the difference between samples significant? (experimental variability)
Chi2 test
dna.methylation.Pnpla5.AL<-
dna.methylation.Pnpla5[dna.methylation.Pnpla5$Group=="AL",]
chisq.test(dna.methylation.Pnpla5.AL[,4:5], correct=F)
3x2 contingency table
Unmeth
Meth
AL-1174
287
254
AL-1180
121 97
AL-1220
163
187

24. Exercise 16: dna_methylation_format2.csv - Answers
Question 1: is the difference between samples significant? (experimental variability)
## test difference between samples for each gene/condition ###
1 dna.methylation.Pno1.AL<-dna.methylation.Pno1[dna.methylation.Pno1$Group=="AL",]
chisq.test(dna.methylation.Pno1.AL[,4:5], correct=F)
2 dna.methylation.Pno1.DR<-dna.methylation.Pno1[dna.methylation.Pno1$Group=="DR",]
chisq.test(dna.methylation.Pno1.DR[,4:5], correct=F)
3 dna.methylation.Pnpla5.AL<-dna.methylation.Pnpla5[dna.methylation.Pnpla5$Group=="AL",]
chisq.test(dna.methylation.Pnpla5.AL[,4:5], correct=F)
4 dna.methylation.Pnpla5.DR<-dna.methylation.Pnpla5[dna.methylation.Pnpla5$Group=="DR",] 1 2 3 4

25. Exercise 16: dna_methylation_format2.csv - Answers
Question 2: is the difference between conditions significant accounting for the variation between samples?
Data need to be in the long format (1 file per gene)
Restructure dna.methylation.Pno1 and dna.methylation.Pnpla5
Rename the columns “Methylation” and “Counts”
Extra credits if you remove the first and the last columns 

26. Exercise 16: dna_methylation_format2.csv - Answers
## prepare file (long format)##
dna.methylation.Pno1.long <- melt(dna.methylation.Pno1[,2:5], ID=c("Sample", "Group"))
dna.methylation.Pnpla5.long <- melt(dna.methylation.Pnpla5[,2:5], ID=c("Sample", "Group"))
colnames(dna.methylation.Pno1.long)[3:4]<-c("Methylation", "Counts")
colnames(dna.methylation.Pnpla5.long)[3:4]<-c("Methylation", "Counts")
head(dna.methylation.Pno1.long)

27. Logistic regression dna_methylation_format2.csv y = β0 + β1*x
Question 2: is the difference between conditions significant accounting for the variation between samples?
Logistic regression: Equivalent to a 2-way ANOVA or a linear model but with a binary outcome
Family: Linear Model (General Linear Model or Generalized Linear Model)
y = f(x)
One factor: y = Constant + Coefficient * x
Two factors: y = Constant + Coefficient1 * x1 + Coefficient2 * x2
y = f(x1, x2 … xn)
In R: Outcome ~ factor1 + factor2
With the interaction: Outcome ~ factor1 + factor2 + factor1*factor2
y = β0 + β1*x

28. Logistic regression dna_methylation_format2.csv
Question 2: is the difference between conditions significant accounting for the variation between samples?
Logistic regression: Equivalent to a 2-way ANOVA but with a binary outcome
In R:
Continuous outcome:
lm or aov(Outcome ~ factor1 + factor2 + factor1*factor2) = glm(Outcome ~ factor1 + factor2 + factor1*factor2)
Binary outcome:
lm or aov(does not work) = glm(Outcome ~ factor1 + factor2 + factor1*factor2, family = binomial())
Data need to be in the long format (1 file per gene)

29. Logistic regression dna_methylation_format2.csv
Question 2: is the difference between conditions significant accounting for the variation between samples?
dna.model.Pno1 <- glm(Methylation ~ Group+Sample, data=dna.methylation.Pno1.long, family = binomial(), weights=Counts)
Group
Methylation AL 1 DR
Group
Methylation
Counts AL
Unmeth 2
Meth 3 DR 1 4

30. Logistic regression dna_methylation_format2.csv
Question 2: is the difference between conditions significant accounting for the variation between samples?
dna.model.Pno1 <- glm(Methylation ~ Group+Sample, data=dna.methylation.Pno1.long, family = binomial(), weights=Counts)
summary(dna.model.Pno1)

31. Logistic regression dna_methylation_format2.csv
Question 2: is the difference between conditions significant accounting for the variation between samples?
dna.model.Pnpla5<-glm(Methylation ~ Group+Sample,data=dna.methylation.Pnpla5.long, family = binomial(), weights=Counts)
summary(dna.model.Pnpla5)

32. Real Example: DNA Methylation
Measures per gene
Chi-Square p<0.001 (FDR)

33. Our addresses: